Question

For an even integer \(n\), prove that \(C(n, n / 2)\) does not have polynomial order. Interpret this in terms of implementing an algorithm that examines this number of subsets of a set.

Short answer

\( C(n, n/2) \) has exponential growth, and examining its subsets is not polynomial time feasible.

Step-by-step solution

Understand the Binomial Coefficient

The binomial coefficient \( C(n, n/2) \), also written as \( \binom{n}{n/2} \), represents the number of ways to choose \( n/2 \) elements from a set of \( n \) elements. It is calculated using the formula: \[ \binom{n}{n/2} = \frac{n!}{(n/2)! (n/2)!} \]

Approximate the Binomial Coefficient

Using Stirling's approximation, \( n! \approx \sqrt{2 \pi n} \left( \frac{n}{e} \right)^n \), one can approximate \( \binom{n}{n/2} \). Applying this, we get:\[\binom{n}{n/2} \approx \frac{2^n}{\sqrt{\pi n/2}}\] This shows that \( \binom{n}{n/2} \) grows faster than any polynomial in \( n \).

Analyze Polynomial Growth

A polynomial of degree \( k \) is expressed as \( p(n) = a_k n^k + a_{k-1} n^{k-1} + \ldots + a_0 \). For a binomial coefficient to have polynomial order, it should not grow faster than \( n^k \) for some \( k \).

Conclude Growth Comparison

Since \( \binom{n}{n/2} \approx \frac{2^n}{\sqrt{\pi n/2}} \) grows exponentially (because of the \( 2^n \) term), it clearly grows faster than any polynomial order (like \( n^k \)).

Interpret in Algorithmic Context

When you implement an algorithm that examines \( \binom{n}{n/2} \) subsets, the complexity is exponential due to the term \( 2^n \). This indicates that the algorithm will not perform efficiently (polynomial time) as \( n \) grows.

Key concepts

Polynomial Growth

Polynomial growth describes a scenario where a function grows in proportion to a fixed power of a variable. For instance, a polynomial function of degree \( k \) can be expressed as:

• \( p(n) = a_k n^k + a_{k-1} n^{k-1} + \ldots + a_0 \)

This means that its growth rate depends on \( n^k \), and as \( n \) increases, the dominant term \( a_k n^k \) dictates the behavior of the function.
Understanding polynomial growth is crucial when analyzing algorithm efficiency. If an algorithm's complexity can be bounded by a polynomial function, it is generally regarded as efficient and feasible for larger inputs. However, when growth is not polynomial, it can signify increasing computational demands that become impractical for large values of \( n \).
This is why the problem, where the binomial coefficient \( \binom{n}{n/2} \) exceeds polynomial growth, highlights the inefficiency of algorithms attempting to handle such values straightforwardly. Such functions grow faster than any polynomial, indicating that they are non-polynomial or super-polynomial.

Stirling's Approximation

Stirling's Approximation is an essential tool for approximating the factorial of a large number. It provides a simplified method to estimate \( n! \), given by:

• \( n! \approx \sqrt{2 \pi n} \left( \frac{n}{e} \right)^n \)

This approximation is particularly useful in complex calculations involving factorials, where direct computation is infeasible due to the sheer size of the numbers involved.
In the context of binomial coefficients like \( \binom{n}{n/2} \), Stirling's Approximation can be used to illustrate exponential growth patterns.By approximating \( n! \) in the formula, we arrive at:

• \( \binom{n}{n/2} \approx \frac{2^n}{\sqrt{\pi n/2}} \)

This highlights not only the growth rate but also that the factor \( 2^n \) dominates, clearly exceeding any polynomial terms, thereby suggesting exponential growth.Stirling's Approximation is widely used for simplifying analyses in probability, statistics, and algorithm design, particularly when dealing with large datasets or combinatorial problems.

Exponential Growth

Exponential growth refers to an increase that becomes ever more rapid in proportion to the growing total number or size. Specifically, a function exhibits exponential growth if it can be described by an expression like \( a b^n \) where \( a \) is a constant and \( b > 1 \). This type of growth surpasses polynomial growth significantly.
In our problem, the binomial coefficient \( \binom{n}{n/2} \), as approximated using Stirling's method, reveals an exponential growth pattern given by:

• \( \approx \frac{2^n}{\sqrt{\pi n/2}} \)

Here, the term \( 2^n \) signifies the exponential nature since it increases exponentially with the increment of \( n \).
Exponential growth in algorithmic contexts is critical because it indicates that as the input size \( n \) rises, the computational requirements or running time of the algorithm increases rapidly. Consequently, algorithms that involve counting problems like binomial coefficients often face challenges in efficiently dealing with large \( n \) due to the exponential growth of the number of operations. Understanding these growth trends is pivotal for designing scalable and practical algorithms.